Quadric surfaces problem please help.

Consider the surfaces x^2+2y^2-z^2+3x=1 and 2x^2+4y^2-2z^2-5y=0.
a. What is the name of each surface?
b. Find an equation for the plane which contains the intersection of these two surfaces.

That is the question. I got part "a" and I am pretty sure it is right. What I really have no idea how to do is "b". Can someone tell me the answer to this and please explain to me how to do it? Thanks in advance.

where is a normal vector to the plane. That means I had to use four points along the yellow curves above to calculate the normal vector (three equations in three unknowns and one point for the (x,y,z) values as well. Probably an easier way though. This is tough for me.

So you're saying I should use the points (1/3,0,1/3), (1/3,0,-1/3), and (0,-2,sqrt(7)) as the three points, right? Just want to make sure, because when I used them the normal vector's k component came out to be 0. Just seemed a little weird to me, so I just wanted to make sure I understood you correctly.